还有一道数列的题已知数列{an}和{bn}都是等差数列,数列{an}的前n项和记作Sn,数列{bn}的前n项和记作Tn,

Sn=n(a1+an)/2 Tn=n(b1+bn)/2
Sn/Tn=(a1+an)/(b1+bn)=2n-3/4n-3
⑴:a2000=1/2(a1+a3999)
b2000=1/2(b1+b3999)
所以,a2000/b2000=(a1+a3999)/(b1+b3999)=(2×3999-3)/(4×3999-3)=8995/15993
⑵：a1999+a1=2a1000
a1900+a100=2a1000
所以,原式=(b100+b1900)/2a1000
=2b1000/2a1000=(b1+b1999)/(a1+a1999)=(4×1999-3)/(2×1999-3)=7993/1995